\(\def\N{\mathbb{N}}\) \(\def\Z{\mathbb{Z}}\) \(\def\Q{\mathbb{Q}}\) \(\def\R{\mathbb{R}}\) \(\def\C{\mathbb{C}}\) \(\def\H{\mathbb{H}}\) \(\def\6{\partial}\) \(\DeclareMathOperator\Res{Res}\) \(\DeclareMathOperator\M{M}\) \(\DeclareMathOperator\ord{ord}\) \(\DeclareMathOperator\const{const}\) \(\DeclareMathOperator{\arccosh}{arccosh}\) \(\DeclareMathOperator{\arcsinh}{arcsinh}\) \(\DeclareMathOperator\id{id}\) \(\DeclareMathOperator\rk{rk}\) \(\DeclareMathOperator\tr{tr}\) \(\def\pt{\mathrm{pt}}\) \(\DeclareMathOperator\colim{colim}\) \(\DeclareMathOperator\Hom{Hom}\) \(\DeclareMathOperator\End{End}\) \(\DeclareMathOperator\Aut{Aut}\) \(\let\Im\relax\DeclareMathOperator\Im{Im}\) \(\let\Re\relax\DeclareMathOperator\Re{Re}\) \(\DeclareMathOperator\Ker{Ker}\) \(\DeclareMathOperator\Coker{Coker}\) \(\DeclareMathOperator\Map{Map}\) \(\def\GL{\mathrm{GL}}\) \(\def\SL{\mathrm{SL}}\) \(\def\O{\mathrm{O}}\) \(\def\SO{\mathrm{SO}}\) \(\def\Spin{\mathrm{Spin}}\) \(\def\U{\mathrm{U}}\) \(\def\SU{\mathrm{SU}}\) \(\def\g{{\mathfrak g}}\) \(\def\h{{\mathfrak h}}\) \(\def\gl{{\mathfrak{gl}}}\) \(\def\sl{{\mathfrak{sl}}}\) \(\def\sp{{\mathfrak{sp}}}\) \(\def\so{{\mathfrak{so}}}\) \(\def\spin{{\mathfrak{spin}}}\) \(\def\u{{\mathfrak u}}\) \(\def\su{{\mathfrak{su}}}\) \(\def\cA{\mathcal{A}}\) \(\def\cB{\mathcal{B}}\) \(\def\cC{\mathcal{C}}\) \(\def\cD{\mathcal{D}}\) \(\def\cE{\mathcal{E}}\) \(\def\cF{\mathcal{F}}\) \(\def\cG{\mathcal{G}}\) \(\def\cH{\mathcal{H}}\) \(\def\cI{\mathcal{I}}\) \(\def\cJ{\mathcal{J}}\) \(\def\cK{\mathcal{K}}\) \(\def\cL{\mathcal{L}}\) \(\def\cM{\mathcal{M}}\) \(\def\cN{\mathcal{N}}\) \(\def\cO{\mathcal{O}}\) \(\def\cP{\mathcal{P}}\) \(\def\cQ{\mathcal{Q}}\) \(\def\cR{\mathcal{R}}\) \(\def\cS{\mathcal{S}}\) \(\def\cT{\mathcal{T}}\) \(\def\cU{\mathcal{U}}\) \(\def\cV{\mathcal{V}}\) \(\def\cW{\mathcal{W}}\) \(\def\cX{\mathcal{X}}\) \(\def\cY{\mathcal{Y}}\) \(\def\cZ{\mathcal{Z}}\) \(\def\al{\alpha}\) \(\def\be{\beta}\) \(\def\ga{\gamma}\) \(\def\de{\delta}\) \(\def\ep{\epsilon}\) \(\def\ze{\zeta}\) \(\def\th{\theta}\) \(\def\io{\iota}\) \(\def\ka{\kappa}\) \(\def\la{\lambda}\) \(\def\si{\sigma}\) \(\def\up{\upsilon}\) \(\def\vp{\varphi}\) \(\def\om{\omega}\) \(\def\De{\Delta}\) \(\def\Ka{{\rm K}}\) \(\def\La{\Lambda}\) \(\def\Om{\Omega}\) \(\def\Ga{\Gamma}\) \(\def\Si{\Sigma}\) \(\def\Th{\Theta}\) \(\def\Up{\Upsilon}\) \(\def\Chi{{\rm X}}\) \(\def\Tau{{T}}\) \(\def\Nu{{\rm N}}\) \(\def\op{\oplus}\) \(\def\ot{\otimes}\) \(\def\t{\times}\) \(\def\bt{\boxtimes}\) \(\def\bu{\bullet}\) \(\def\iy{\infty}\) \(\def\longra{\longrightarrow}\) \(\def\an#1{\langle #1 \rangle}\) \(\def\ban#1{\bigl\langle #1 \bigr\rangle}\) \(\def\llbracket{{\normalsize\unicode{x27E6}}} \def\rrbracket{{\normalsize\unicode{x27E7}}} \) \(\def\lb{\llbracket}\) \(\def\rb{\rrbracket}\) \(\def\ul{\underline}\) \(\def\ol{\overline}\)

10  Residue theorem

Definition 10.1 Let \(z_0\) be an isolated singularity of a holomorphic function \(f\colon U\to\C.\) The residue \(\Res_{z_0}(f)\) is the residue of the Laurent series Equation 9.8. In other words, \[f(z)=\sum_{n=-\iy}^\iy a_n(z-z_0)^n\implies \Res_{z_0}(f)=a_{-1}.\]

Proposition 10.1 Let \(z_0\) be an isolated singularity of both the holomorphic functions \(f,g\colon U\to\C.\)

  1. \(\Res_{z_0}(f+g)=\Res_{z_0}(f)+\Res_{z_0}(g)\)
  2. \(\Res_{z_0}(\la f)=\la\Res_{z_0}(f)\) for all \(\la\in\C\)
  3. If \(z_0\) is a pole of order \(m,\) then \(\Res_{z_0}(f)=\frac{1}{(m-1)!}\lim_{z\to z_0}h^{(m-1)}(z)\) for \(h(z)=(z-z_0)^mf(z).\)

Proof.

  1. and b.are left as an Exercise.

  2. If \(f=\sum_{n=-m}^\iy a_n(z-z_0)^n,\) then \(h(z)=\sum_{n=-m}^\iy a_n(z-z_0)^{n+m}\) is a power series which we can differentiate termwise by Theorem 5.2 to get \[h^{(m-1)}(z)=\sum_{n=-m}^\iy (n+m)(n+m-1)\cdots(n+2) a_n(z-z_0)^{n+1}.\]

The limit \(z\to z_0\) of this power series is its constant term, \((m-1)!a_{-1}.\)

Example 10.1  

\(\Res_0(z^n)=\begin{cases} 0 & \text{if $n\neq-1$},\\ 1 & \text{if $n=-1$}. \end{cases}\)

Example 10.2  

The best way to compute a residue is often to work out the Laurent expansion using the calculus of formal power series from Chapter 5.

  1. Consider \(f(z)=\frac{\sin(z^2)}{z^7}.\) As \(\sin(z)=\sum_{n=0}^\iy\frac{(-1)^n}{(2n+1)!}z^{2n+1},\) we have \[f(z)=\sum_{n=0}^\iy\frac{(-1)^n}{(2n+1)!}z^{4n+2-7},\] hence \(\Res_0f(z)=-1/6\) is the coefficient of the term with \(n=1.\)

  2. Let \(g(z)=\frac{\cos(z)-1}{(\exp(z)-1)^3}.\) It suffices to work with finite portions of the involved power series. For this, introduce the notation \(O(z^m)\) to mean ‘terms involving \(z^n\) of order \(m\leqslant n\)’. We will use this in an informal way, which can be made precise in algebra by saying that \(O(z^n)\) is the ideal in the ring of formal power series generated by \(z^n.\) Notice the properties \[\begin{align} O(z^m)O(z^n)&=O(z^{m+n}),\\ O(z^m)^n&=O(z^{mn}),\\ O(z^m)+O(z^n)&=O\bigl(z^{\min(m,n)}\bigr),\\ 1/O(z^n)&=O(z^{-n}). \end{align}\] Now compute \[\begin{align*} \exp(z)-1&=z+z^2/2+O(z^3),\\ \implies(\exp(z)-1)^3&=\frac{z^6}{8}+\frac{3 z^5}{4}+\frac{3 z^4}{2}+z^3+O(z^5)\\ &=z^3+\frac{3z^4}{2}+O(z^5). \end{align*}\] To find the expansion of the inverse \(1/(\exp(z)-1)^3,\) we apply the proof strategy of Proposition 5.1(b) to \(z^{-3}(\exp(z)-1)^3=1+\frac{3z}{2}+O(z^2).\) In other words, we compare coefficients to solve \[\left(1+\frac{3z}{2}+\ldots\right)(a_0+a_1z+a_2z^2+\ldots)=1\]

    recursively for \(a_0, a_1, a_2.\) This gives \[1/(\exp(z)-1)^3=z^{-3}\left(1-\frac{3z}{2}+\frac{9z^2}{4}+O(z^3)\right)=z^{-3}-\frac{3z^{-2}}{2}+\frac{9z^{-1}}{4}+O(z^0).\]

    Finally, multiply this by \(\cos(z)-1=-\frac{z^2}{2}+\frac{z^4}{24}+O(z^5)\) to get \[g(z)=-\frac{1}{2 z}+\frac{3}{4}-\frac{13 z}{12}-\frac{z^2}{16}+O(z^2)\]

    from which we read off \(\Res_0g(z)=-1/2.\)

Definition 10.2 Let \(\ga\colon[a,b]\to\C\) be a closed piecewise C1 curve and let \(z_0\in\C\setminus\ga([a,b]).\) The winding number of \(\ga\) around \(z_0\) is\[W_{z_0}(\ga)=\frac{1}{2\pi i}\int_\ga\frac{dz}{z-z_0}.\]

Example 10.3  

Let \(\ga\colon[0,2\pi]\to\C,\) \(\ga(t)=z_0+re^{it},\) be the boundary curve of the disk \(\ol D_r(z_0).\) Then Equation 6.6 implies \(W_{z_0}(\ga)=1.\) More generally, for every \(w_0\in D_r(z_0)\) the Cauchy integral formula Equation 8.1 gives\[W_{w_0}(\ga)=1.\]

Pick a subdivision \(a=t_0<t_1<\cdots<t_n=b\) such that \(\ga|_{[t_{k-1},t_k]}\) is continuously differentiable. Using branches of the logarithm, we can write \(\ga(t)=z_0+r(t)e^{i\th(t)}\) for \(r(t)=|\ga(t)-z_0|\) and a continuous function \(\th\colon[a,b]\to\R\) with \(\th|_{[t_{k-1},t_k]}\) continuously differentiable. With this notation, the following holds.

Proposition 10.2  

  1. We have \(W_{z_0}(\ga)=\frac{\th(b)-\th(a)}{2\pi},\) which is an integer.
  2. For closed curves \(\ga_0,\) \(\ga_1\) in \(\C\setminus\{z_0\}\) we have \[W_{z_0}(\ga_0)=W_{z_0}(\ga_1)\iff\text{$\ga_0,$ $\ga_1$ are homotopic in $\C\setminus\{z_0\}.$}\]

Proof.

  1. Over \([t_{k-1},t_k]\) we have \(\ga'(t)=r'(t)e^{i\th(t)}+r(t)\th'(t)e^{i\th(t)}.\) Using\[\frac{\ga'(t)}{\ga(t)-z_0}=\frac{r'(t)}{r(t)}+i\th'(t) \tag{10.1}\] we compute \[2\pi i W_{z_0}(\ga)=\sum_{k=1}^n\int_{t_{k-1}}^{t_k}\frac{\ga'(t)}{\ga(t)}dt\] by Equation 10.1 we have \[\sum_{k=1}^n\int_{t_{k-1}}^{t_k}\frac{\ga'(t)}{\ga(t)}dt = \sum_{k=1}^n\left(\int_{t_{k-1}}^{t_k}\frac{d}{dt}\log r(t)dt + i\int_{t_{k-1}}^{t_k}\th'(t)dt\right)\] using FTC, the right hand side becomes \[\sum_{k=1}^n \Bigl(\log r(t_k)- \log r(t_{k-1})+i\th(t_k)-i\th(t_{k-1})\Bigr)\] observing that these are telescoping sums, we conclude that \[2\pi i W_{z_0}(\ga)= i\bigl(\th(t_n)-\th(t_0)\bigr)=i\bigl(\th(b)-\th(a)\bigr)\]and the claimed formula follows. From \(\ga(a)=\ga(b)\) we have \(e^{i\th(a)}=e^{i\th(b)},\) so Proposition 1.5 implies \(\th(b)-\th(a)=2\pi k\) for some \(k\in\Z.\)

  2. By Theorem 7.2 applied to \(U=\C\setminus\{z_0\}\) and \(f(z)=1/(z-z_0),\) the implication ‘\(\Longleftarrow\)’ holds.

    Conversely, assume \(W_{z_0}(\ga_0)=W_{z_0}(\ga_1).\) Write \(\ga_k(t)=z_0+r_k(t)e^{i\th_k(t)}\) for \(k=0,1.\) As \(\ga_k(a)=\ga_k(b),\) we have \(r_k(a)=r_k(b)\) and \(\th_k(a)=\th_k(b)-2\pi i n_k\) for some \(n_k\in\Z.\) By assumption, \(\frac{\th_0(b)-\th_0(a)}{2\pi}=\frac{\th_1(b)-\th_1(a)}{2\pi},\) hence \(n_0=n_1.\) Define the homotopy by \[\Ga_s(t)=z_0+[(1-s)r_0(t)+sr_1(t)]\exp(i(1-s)\th_0(t)+is\th_1(t)).\]

    Clearly, \(\Ga_k=\ga_k\) for \(k=0,1.\) The only thing to check is that \(\Ga_s\) is a closed curve for each \(s\in[0,1],\) that is, to show \(\Ga_s(b)=\Ga_s(a).\) As \(r_k(a)=r_k(b),\) this is equivalent to \[\exp\bigl(i(1-s)(\th_1(b)-\th_1(a))+is(\th_0(b)-\th_0(a))\bigr)=1.\]

    This holds since \(\th_1(b)-\th_1(a)=\th_0(b)-\th_0(a)=2\pi n_0\) and \[\exp\bigl(i(1-s)2\pi n_0+is2\pi n_0)=\exp(2\pi in_0\bigr)=1.\]

Theorem 10.1 (Residue theorem) Let \(U\subset\C\) be an open set, \(\ga\colon[a,b]\to U\) a null-homotopic piecewise C1 loop in \(U,\) and \(z_1,\ldots, z_n\in U\setminus\ga([a,b])\) points not on the loop \(\ga.\) If \(f\) is a holomorphic function on \(U\setminus\{z_1,\ldots, z_n\},\) then

\[\frac{1}{2\pi i}\int_\ga f(z)dz=\sum_{k=1}^n W_{z_k}(\ga)\Res_{z_k}(f).\]

Proof.

For each \(k=1,\ldots,n\) we find \(r_k>0\) and Laurent expansions \[f(z)=\sum_{n=-\iy}^{\iy}a_n^{(k)}(z-z_k)^n\qquad\text{for all $z\in D^\t_{r_k}(z_k).$}\]

Let \(P_k(z)=\sum_{n=-\iy}^{-1}a_n^{(k)}(z-z_k)^n\) be the corresponding principal part. Hence \(a_{-1}^{(k)}=\Res_{z_k}(f).\) By Proposition 9.1, \(P_k(z)\) is holomorphic on \(D^\t_{+\iy}(z_0).\) Also, \(f(z)-\sum_{k=1}^n P_k(z)\) defines a holomorphic function on \(U\) since its singularities at \(\{z_1,\ldots,z_n\}\) are removable. By Cauchy’s Theorem 7.1, \[\int_\ga f(z)dz=\sum_{k=1}^b\int_\ga P_k(z)dz.\]

By Proposition 10.2(b), the loop \(\ga\) is homotopic in \(\C\setminus\{z_k\}\) to \(\6 D_{r_k}(z_k)\ast\cdots\ast\6 D_{r_k}(z_k)\) (the same loop traversed \(W_{z_k}(\ga)\)-times). Combined with the fact that \(P_k\) is holomorphic on \(\C\setminus\{z_k\},\) this implies by Theorem 7.2 that \[\begin{align*} &\int_\ga P_k(z)dz=W_{z_k}(\ga)\int_{\6D_{r_k}(z_k)} P_k(z)dz \\ &=W_{z_k}(\ga)\int_{\6D_{r_k}(z_k)}\sum_{n=-\iy}^{-1}a_n^{(k)}(z-z_k)^ndz \end{align*}\] by uniform convergence, \[W_{z_k}(\ga)\int_{\6D_{r_k}(z_k)}\sum_{n=-\iy}^{-1}a_n^{(k)}(z-z_k)^ndz = W_{z_k}(\ga)\sum_{n=-\iy}^{-1}a_n^{(k)}\int_{\6D_{r_k}(z_k)}(z-z_k)^ndz\] using Equation 6.6 we have \[W_{z_k}(\ga)\sum_{n=-\iy}^{-1}a_n^{(k)}\int_{\6D_{r_k}(z_k)}(z-z_k)^ndz = W_{z_k}(\ga)a_{-1}^{(k)}2\pi i\] as required.

10.1 Exercises

Exercise 10.1

Determine the winding numbers \(W_{z_0}(\ga)\) in the following cases.

  1. \(\ga\colon[0,2\pi]\to\C,\) \(\ga(t)=e^{it},\) for \(z_0=0,2.\)
  2. \(\ga\colon[2\pi,10\pi]\to\C,\) \(\ga(t)=\begin{cases} te^{it} & \text{if $t\in[2\pi, 6\pi],$}\\ 12\pi - t & \text{if $t\in[6\pi,10\pi],$} \end{cases}\) for \(z_0=0, 20.\)
  3. \(\ga\colon[0,2\pi]\to\C,\) \(\ga(t)=(2+\cos(t))e^{2it},\) for \(z_0=0,4.\)
Exercise 10.2

Use the residue theorem to compute \[\int_{\partial D_r(0)}\frac{(z+2)^2}{z(z-1)^2}dz\]

for all \(0<r\neq1.\)

Hint: Consider the cases \(r>1\) and \(r<1\) separately.

Exercise 10.3

Apply the residue theorem to the contour \(\partial D_1(0)\) and a suitable holomorphic function \(f(z)\) to compute the integral\[\int_0^{2\pi}\frac{dx}{2+\cos(x)}.\]

Exercise 10.4

Compute the integral \[\int_{0}^{+\iy}\frac{\cos(x)}{\sqrt{x}}dx\] by considering \(f(z)=\frac{e^{iz}}{\sqrt{z}}\) and the following contour \(\ga_{r,R}\) for \(r\to 0,\) \(R\to+\iy.\)

Hint: Recall that \(\int_0^{\iy}e^{-x^2}dx=\sqrt{\pi}/2.\)

Exercise 10.5

Compute the integral \[\int_{-\iy}^{+\iy}\frac{dx}{1+x^6}\]

by applying the residue theorem to \(f(z)=\frac{1}{1+z^6}\) and the following upper semi-circular contour \(\ga_R\) for \(R\to+\iy.\)